Note: The map you propose is not an isomorphism! It’s not even a ring homomorphism.
You send $([1]_3,[1]_4)$ to $[5]_{12}$, but the square of $[5]_{12}$ is $[1]_{12}$, and the square of $([1]_3,[1]_4)$ is itself. So your map does not send products to products. Altenrnatively, you send $([1]_3,[0]_4)$ to $[4]_{12}$. But the product of $([1]_3,[1]_4)$ and $([1]_3,[0]_4)$ is $([1]_3,[0]_4)$, so the product of their images should be $[4]_{12}$, and yet $[5]_{12}\times[4]_{12}=[20]_{12}=[8]_{12}\neq[4]_{12}$.
To prove uniqueness (once you find the right isomorphism), the key is whether $\mathbb{Z}/12\mathbb{Z}$ has nontrivial automorphisms or not. If the only automorphism is the identity, and $\phi,\psi$ are two isomorphisms, then since $\phi\circ\psi^{-1}$ is an automorphism of $\mathbb{Z}/12\mathbb{Z}$ it must equal the identity, hence $\phi=\psi$. Conversely, if $\mathbb{Z}12\mathbb{Z}$ has a non-identity automorphism $\theta$, and $\phi$ is an isomorphism, then $\theta\circ\phi$ is also an isomorphism, but $\theta\phi\neq\phi$ since $\phi$ is invertible, but $\theta\neq\mathrm{id}$.
If you require your ring morphisms to be unital, then any automorphism must send $[1]$ to $[1]$, and hence by induction if $\theta$ is an automorphism then $\theta([n])=[n]$ for all $n$ and the result follows.
If you do not require your ring morphisms to be unital, then since $[1][1]=[1]$, then $\theta([1])$ must be an idempotent. Moreover, because $\theta$ is also a group isomorphism, it must send $[1]$ to an additive generator of $\mathbb{Z}/12\mathbb{Z}$. The additive generators are $[1]$, $[5]$, $[7]$, and $[11]$, and the only idempotent among them is $[1]$ (the rest have $[5][5]=[7][7]=[11][11]=[1]$). So the automorphism must still send $[1]$ to $[1]$ and the result follows as before.
*italic text*to produce italic text, instead of trying to fake it by putting it in math mode. Math italic is different from regular italic, and looks bad. Compare*unique*vs.$unique$: unique vs. $unique$ – Arturo Magidin Jan 28 '21 at 22:50